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Stitz-Zeager_College_Algebra_e-book

# Well just have to use our own good judgment when

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Unformatted text preview: cos2 (t) + 41 = 1. Solving, we x get cos(t) = ± 16x2 −1 16x2 √ =± 16x2 −1 . Since t belongs to − π , 0 2 4√ | |x 16−x2 = 4|x| . (The absolute values here ∪ 0, π , we know 2 cos(t) ≥ 0, so we choose cos(t) are necessary, since x could be negative.) Since the domain of arccsc(x) requires |x| ≥ 1, the domain of arccsc(4x) requires |4x| ≥ 1. Using Theorem 2.3, we can rewrite this as the compound 1 inequality 4x ≤ −1 or 4x ≥ 1. Solving, we get x ≤ − 1 or x ≥ 4 . Since we had no 4 √ additional restrictions on t, the equivalence cos(arccsc(4x)) = 1 1 −∞, − 4 ∪ 4 , ∞ . 16x2 −1 4 |x | holds for all x in 10.6 The Inverse Trigonometric Functions 10.6.2 711 Inverses of Secant and Cosecant: Calculus Friendly Approach π In this subsection, we restrict f (x) = sec(x) to 0, π ∪ π , 32 2 y y 3π 2 1 π π 2 π 3π 2 x −1 π 2 reﬂect across y = x f (x) = sec(x) on 0, π 2 ∪ π, −− − − − −→ −−−−−− 3π 2 switch x and y...
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